a proof involving triangle ABC is shown below what title best describes the proof

Given data:
The given figure of the triangle.
In the triangle ACD and triangle BCD.
[tex]\begin{gathered} AC\cong BC(\text{given)} \\ \angle1\cong\angle2 \\ CD\cong CD(Common) \\ \Delta ACD\cong\Delta\text{BCD(side angle side SAS)} \\ \angle A\cong\angle B(CPCT) \end{gathered}[/tex]Thus, option (B) is correct.